# What exactly does κ stand for in this equation connecting solubility and molar conductivity?

I came across the following equation while studying electrochemistry ("Physical Chemistry" by Wallwork and Grant):

The solubility $s$ of a sparingly soluble salt can be determined from conductivity measurements, provided that the molar conductivities of the ions at the temperature of these measurements are known to enable $\Lambda_0$ to be calculated. Since the saturated solution is dilute, $\Lambda_0 \approx \Lambda = \kappa/c$, where $c = s~(\pu{mol m-3})$. The conductivity $\kappa$ is the conductivity of a saturated solution of the salt in conductance water minus that of the water alone. The solubility $s$ is calculated from: $$s = \kappa/\Lambda_0 \tag{8.36}$$

It's mentioned that κ is the conductivity of the saturated solution of the salt in conductance water, minus that of the water alone. Now the thing is, from what I've read, conductance water has NO (or negligible) conductivity.

In that case, what's the point of having to subtract the conductivity of conductance water? Or is there something I've misunderstood here?

• Let's clear up one thing first. There is no such thing as conductance water. It is a totally weird phrase. I can only assume that pure water was the intended phrase. I'd suspect that the book is a translation, or written by an author in what is his second language. – MaxW Sep 23 '16 at 16:06
• The specific conductance, $\kappa$ (kappa) is the reciprocal of the specific resistance. – MaxW Sep 23 '16 at 16:12
• I'm working on an answer. The author has written this in a slipshod manner. This will never make sense unless the whole thing gets explained using the right terminology. I'm going to lunch. If someone else wants to jump in fine. In the meantime you can look at the wikipedia article on conductivity en.wikipedia.org/wiki/Conductivity_(electrolytic) – MaxW Sep 23 '16 at 16:47

Nomenclature changes and authors don't always follow the rules. The more typical use now is $\Lambda_m$ for the molar conductivity and $\Lambda_m^0$ for the limiting molar conductivity.

The molar conductivity, $\Lambda_m$, is given by the equation

$\Lambda_m = \dfrac{\kappa}{c}\tag{1}$

where:

• $\kappa$ is the specific conductance (which is what you measure...)
• $c$ is the concentration (which you know because of how the solution was made)

Now typically the specific conductance, $\kappa$, is plotted against concentration. For dilute solutions, there will be a linear relationship where the slope of the line is the limiting molar conductivity, $\Lambda_m^0$. However at high concentrations the relationship becomes non-linear and the actual conductance is less than what the linear relationship would predict. That is because of the formation of cation and anion clusters. The local effect is thus to make the cluster uncharged (or charged less is the charges don't balance to zero).

Now the author of your book specified a sparingly soluble salt, so at low concentrations $\Lambda_m \approx \Lambda_m^0$, and we can substitute for $\Lambda_m$ to get:

$\Lambda_m^0 = \dfrac{\kappa}{c}\tag{2}$

which can be rearranged to give:

$c = \dfrac{\kappa}{\Lambda_m^0}\tag{3}$

If the conductivity of the solution is much greater than that of pure water then this is fine. The measured specific conductance will be directly proportional to the concentration since $1/\Lambda_m^0$ is a constant.

However the author is talking about a salt of limited solubility, so the conductivity of pure water is not negligible. Going back to the plot of specific conductance, $\kappa$, against concentration, $c$, we can determine that at 0 concentration there is some conductance which I'll call $\kappa_w$ which is the conductance of pure water. So for a salt with limited solubility the equation is:

$c = \dfrac{\kappa - \kappa_w}{\Lambda_m^0}\tag{4}$

If you look back at equation (2) you can see why equation (3) doesn't work for solutions of very low conductivity. When $c=0$ equation (2) blows up since you can't divide by $0$.

So the author used $\kappa$ in two different ways which is very poor writing. We could define $\kappa'= \kappa - \kappa_w\tag{5}$ then $c = \dfrac{\kappa'}{\Lambda_m^0}\tag{6}$

The conductivity $\kappa$ of pure water is very low. The value for highly purified water at a temperature of $T=25\ \mathrm{^\circ C}$ is only $\kappa=5.50\times10^{-6}\ \mathrm{S\ m^{-1}}$. However, the conductivity of water is sensitive to dissolved impurities. The conductivity of water that is saturated with atmospheric $\ce{CO2}$ is already increased to $\kappa=1.10\times10^{-4}\ \mathrm{S\ m^{-1}}$. Therefore, if necessary, the actual conductivity of the water that is used in the experiment should be measured and subtracted from the measured values for the concerned salt solutions.

Certainly, when measuring the conductivity of strong electrolyte solutions at high concentrations, the contribution of water is not significant and may be neglected. For example, the conductivity of a potassium chloride solution with a molality of $b=1.0\ \mathrm{mol\ kg^{-1}}$ is about $\kappa\approx10.9\ \mathrm{S\ m^{-1}}$.

However, if you want to measure the conductivity of ions in solutions of sparingly soluble salts (for example in order to calculate the solubility of such salts as mentioned in your book), the contribution of water can be significant. For example, the solubility product constant for silver chloride $(\ce{AgCl})$ is $K_\mathrm{sp}=1.77\times10^{-10}$. Ignoring activity coefficients, this solubility corresponds to a concentration of about $c=1.3\times10^{-5}\ \mathrm{mol\ l^{-1}}$ for $\ce{Ag+}$ as well as $\ce{Cl-}$. Note that this value is not much larger than the concentration of $\ce{H+}$ and $\ce{OH-}$ in pure water, and that the ions $\ce{H+}$ and $\ce{OH-}$ have a relatively high molar conductivity $\lambda$.